Flag General Physics> find the average frictional force needed ...
question mark

find the average frictional force needed to accelerate a car weighing 500 kg from rest to 72 km/h in a distance of 25 m.

prasanjeet kumar , 11 Years ago
Grade 8
anser 1 Answers
Saurabh Koranglekar

Last Activity: 5 Years ago

To find the average frictional force needed to accelerate a car from rest to a certain speed, we can break the problem down into several logical steps. First, we need to convert the speed from kilometers per hour to meters per second, then we can calculate the car's acceleration, and finally, we can use Newton's second law to find the average frictional force.

Step 1: Converting Speed

The speed of 72 km/h needs to be converted to meters per second. The conversion factor is that 1 km/h is equal to approximately 0.27778 m/s. Therefore:

  • 72 km/h * 0.27778 m/s/km/h = 20 m/s

Step 2: Calculating Acceleration

Next, we need to calculate the acceleration required to reach this speed over a distance of 25 meters. We can use the kinematic equation:

v² = u² + 2as

Where:

  • v = final velocity (20 m/s)
  • u = initial velocity (0 m/s, since the car starts from rest)
  • a = acceleration (what we are trying to find)
  • s = distance (25 m)

Plugging in the values, we have:

20² = 0² + 2a(25)

This simplifies to:

400 = 50a

Solving for 'a', we find:

a = 8 m/s²

Step 3: Finding the Average Frictional Force

Now that we know the acceleration, we can find the average frictional force needed to achieve this acceleration using Newton's second law:

F = ma

Where:

  • F = force (frictional force in this case)
  • m = mass of the car (500 kg)
  • a = acceleration (8 m/s²)

Plugging in the values gives us:

F = 500 kg * 8 m/s² = 4000 N

Final Thoughts

The average frictional force required to accelerate the car weighing 500 kg from rest to 72 km/h over a distance of 25 meters is 4000 Newtons. This force is what needs to be exerted through friction between the tires and the road to achieve the desired acceleration.

star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments